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Does anyone know how to solve $a_{n+1}=1-Ce^{-a_n}$ explicitly for $a_n$ in terms of $n$ and $a_0$, where $C$ is constant?

SAM
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  • What is the value of $a_0$? (and hence $a_1$?) –  Jun 30 '16 at 08:28
  • Its general. But you can suppose $a_0=0$. – SAM Jun 30 '16 at 08:38
  • Thanks, If $a_0=0$ then $a_1=1-C$ and $a_2=1-Ce^{c-1}$. This leads to $a_3=1-Ce^{(1-Ce^{c-1})}$. This may be a tough thing to solve for general $a_n$. –  Jun 30 '16 at 08:42
  • Thanks. Those values are correct and simple, but I'm looking for some explicit function. It can be an approximate explicit function. – SAM Jun 30 '16 at 09:20
  • Ok, I have dumped my thoughts, you may find it useful, I can only think about approaching this in terms of limits of the sequence and approximate results. Thanks. –  Jun 30 '16 at 09:45

2 Answers2

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I think I correct in writing \begin{equation} e^{-a_n} < 1-a_n+\frac{a_{n}^{2}}{2} \end{equation} For the purposes of ease, let $C=1$. Then \begin{align} 1-e^{-a_{n}} &< 1-\left(1-a_n \frac{a_n^{2}}{2}\right) \\ &= a_n-\frac{a_n^{2}}{2} \\ &=a_n\left(1-\frac{a_n}{2}\right) \end{align} Therefore \begin{align} a_n+1 &< a_n\left(1-\frac{a_n}{2}\right) \\ \implies \frac{a_n+1}{a_n} &< 1-\frac{a_n}{2} \end{align}

I wonder if at all these musings aid you?

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Long story short, you won't have the explicit solution, but a decent approximation may be possible.

Now, as $C<1$, there is a limit value which can be expressed in terms of Lambert W function (specifically, $a_\infty=1+W(-{C\over e})$, see the picture below, that's $a_\infty$ as a function of $C$), and $a_n$ tends to that limit exponentially, i.e. $a_n\approx a_\infty+c_1\cdot c_2^{-n}$, where $c_1$ and $c_2$ are some constants. enter image description here

As $C=1$, the limit is $0$, and $a_n$ tends to it in a slower manner, namely, as $a_n\approx{c_1\over n}$.

As $C>1$, there is no limit. $a_n$ tends to negative infinity, and pretty fast at that.

Ivan Neretin
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